# Harm osc, anharmonic perturbation

## Homework Statement

Particle bound by
$$V = \frac{1}{2} m \omega^2 x^2 - a x^3$$
for small x. Show that the mean position of the particle changes with the energy of the eigenstates when $$a$$ is small, so first order perturbation theory works.

## Homework Equations

For the harmonic oscillator
$$x = \sqrt{\frac{\hbar}{2m\omega}}(a^{\dagger}+a)$$

## The Attempt at a Solution

That x^3 perturbation will give an odd number of creation/destruction operators, so there's no shift in energy eigenvalues to first order in perturbation theory. But how does that help answering the question?

Dr Transport
Gold Member
There is a first order transition, but it is between different energy states.

But how does that help answering the question?

Noone have any ideas on this?

I'm not going to give up on this! The question asks to calculate the mean position of the particle, so if I do that:
$$<x> = \sqrt{\frac{\hbar}{2m\omega}}<n|a^{\dagger}+a|n> = 0$$
That's just zero, haveing nothing to do with energy eigernstates or size of a?

Sure I could calculate the perturbated energy to second order, getting terms like
$$a^2 \big( \frac{\hbar}{2m\omega} \big)^3 <n|a^{\dagger} aa \frac{1}{E_n-H_0} a^{\dagger} a a^{\dagger} |n> = -n(n+1)^2 \frac{\hbar^2 a^2}{8m^3 \omega^4}$$
and so on, but I don't see how that has anyhting to do with the question asked.

Shouldn't you consider the expanded eigenkets?

$$| n > = |n^{(0)}>+\lambda |n^{(1)}> + \lambda^2 |n^{(2)}> + \dots$$

Then you get to first order something like:

$$<x> = < n^{(0)} | x | n^{(0)} > + \lambda \left( < n^{(0)} | x | n^{(1)} > + < n^{(1)} | x | n^{(0)} > \right)+ \mathcal{O} (\lambda^2) = \lambda \left( < n^{(0)} | x | n^{(1)} > + < n^{(1)} | x | n^{(0)} > \right)+ \mathcal{O} (\lambda^2)$$

Could that be the correct approach?

Last edited:
I think that you dont really have to solve for the corrected energy's. You need the new eigenkets, and then just solve for the expectation value, <x>, by taking the inner product of x and the eigenkets.